Solve x^3 - 245.711x^2 + 2.764156x + 0.057072 = 0.

Step-by-step solution

1. 1

   Predict the scale of the roots before computing. The $x^{2}$ coefficient is enormous compared with the others, so one root must be close to $245.711$ (it alone almost balances the cubic term), and the remaining two must be tiny, roughly the roots of the near-quadratic $-245.711x^{2}+2.764156x+0.057072\approx 0$. Expecting a $10^{4}$ spread in magnitude tells you which checks will and will not be sensitive.

2. 2

   Locate the large root. For $x$ near $245.711$ the cubic is dominated by $x^{2}(x-245.711)$, and Newton iteration from $x_{0}=245.7$ converges immediately to

   $x_{3}=245.6997489169$

3. 3

   Locate the two small roots by sign changes. Evaluating $f$ near zero: $f(-0.011)<0$, $f(0)=0.057072>0$, $f(0.022)<0$, so there is a root in $(-0.011,0)$ and another in $(0,0.022)$. Newton iteration in 50-digit arithmetic gives

   $x_{1}=-0.0106203880,\qquad x_{2}=0.0218714711$

4. 4

   Verify by substitution, not by Vieta alone. This is the crucial safeguard. The sum of roots must be $245.711$ and the product $-0.057072$, but the large root dominates the sum so completely that wrong small roots still appear to pass: the incorrect pair $-0.0150152414,\ 0.0262657963$ satisfies the Vieta sum check to four decimals (its product, $-0.09690$, misses badly), yet substituting either into $f$ gives $-0.0398$, not $0$.

5. 5

   Confirm each root directly. Substituting the correct values gives $f(x_{1})$, $f(x_{2})$ and $f(x_{3})$ all below $10^{-40}$ in 50-digit arithmetic ✓. Only substitution into the original polynomial distinguishes a genuine root from a plausible near-miss.

6. 6

   State the three roots.

   $x\approx-0.0106203880,\qquad x\approx 0.0218714711,\qquad x\approx 245.6997489169$

   All three are real, which matches the discriminant being positive; the cubic crosses the axis three times, twice within a hundredth of the origin.